Question

Describe an algorithm that finds the Cantor expansion of an integer.

Short answer

$$\begin{gathered} 1:n=1 \\ 2:\text{while(num}\le n!) \\ 3:n=n+1 \\ \text{4:end while} \\ 5:n=n-1 \\ 6:\text{while}(\text{num}>0) \\ \text{7:Let a be the integral part of}num/(n!) \\ 8:{\text{Print}}^{\mathrm{\prime }}a(n!) \\ \text{9:}num=num-(a\times n) \\ \text{10:}n=n-1 \\ \text{11:end while} \end{gathered}$$

Step-by-step solution

Step 1:

The algorithm first finds the smallest n for which $num>n!$and then subtract one from it

The is much like finding the number of digits in a number $num$while wanting to write its decimal expansion is d digits long then the expansion starts with$ax{10}^4$for some coefficient a because $d+1$is the smallest number $x$for which the$num>{10}^x$

Step 2:

The algorithm then finds the coefficients for each of the terms

This is done in the same way as when we find coefficient. the coefficients in the decimal expansion. the coefficient of${10}^d$for some$d$when we find the decimal expansion for$num$is the integral part of$num/\left({10}^d\right)$when$num$is d digits long. to ensure that at each step. Remain$d$digits long. we update num by chopping off from it the part that we have already found. here also we do the same